When galvanometer of unknown resistance connected across a series combination of two identical batteries each of 1.5 V, the current through the resistor is 1A. When it is connected across parallel combination of the same batteries, the current through it is 0.6 A. The internal resistance of each battery is

Options

A. `1//5Omega`
B. `1//4Omega`
C. `1//3 Omega`
D. `1//2Omega`

Answer

C

Explanation

`I_(S)=(E_(1)+E_(2))/(R+r_(1)+r_(2))`
`:.1=(1.5+1.5)/(R+2r)`
`R+2r=3` ....(i)
`I_(P)=(E)/(R+(r)/(n))`
`0.6=(E)/(R+r/2)`
`0.6=(2E)/(2R+r)`
`0.6=(2xx1.5)/(2R+r)`
`:.1.2R+0.6r=3` ....(ii)
By solving equation (i) and (ii), we have
`r=(1)/(3) Omega`.